Let $\Omega$ be a measure space with measure $\lambda$.
Denote the space of simple functions by: $$\mathcal{S}:=\{s:\Omega\to\mathbb{C}:s=\sum_{k=1}^{K<\infty}s_k\chi_{A_k:\lambda(A_k)<\infty}\}$$
Denote the positive and negative part of the real and imaginary part by: $$f=\Re_+f-\Re_-f+i\Im_+f-i\Im_-f=:\sum_{\alpha=0\ldots3}i^\alpha f_\alpha$$
Define for positive functions: $$\int fd\lambda:=\sup_{s\in\mathcal{S}:s\leq f}\int sd\lambda\quad(f\geq 0)$$ and for complex functions: $$\int fd\lambda:=\sum_{\alpha=1\ldots3}i^\alpha\int f_\alpha d\alpha$$ as long as all terms of the sum are finite.
(Note, measurability is not required here!)
Then a direct consequence is: $$|\int fd\lambda|\leq\int|f|d\lambda$$ which shows that the a suitable integrability condition will be: $$\int|f|d\lambda<\infty\implies\int f_\alpha d\lambda<\infty\quad(\alpha=0\ldots3)$$
Now, is it possible for the converse to actually fail: $$\int|f|d\lambda<\infty\impliedby\int f_\alpha d\lambda<\infty\quad(\alpha=0\ldots3)$$ and what would be a demonstrative example?
Exploiting the estimates: $$f_\alpha\leq|f|$$ $$|f|\leq\sum_{\alpha=0\ldots3}f_\alpha$$ one has by additivity and monotonicity of the integral: $$\int f_\alpha d\lambda\leq\int|f|d\lambda$$ $$\int|f|d\lambda\leq\sum_{\alpha=0\ldots3}\int f_\alpha d\lambda$$
Concluding that: $$f\in L(\lambda)\iff f_\alpha\in L(\lambda)\quad(\alpha=0\ldots3)$$